For the reaction $A \rightarrow B$ the following graph was obtained. The time required (in seconds) for the concentration of $A$ to reduce to $2.5 g L^{-1}$ (if the initial concentration of A was $50 g L^{-1}$ ) is (Nearest integer)
Given : $\log 2=0.3010$
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Solution
As it is difficult to predict order using data provided in graph.
For specific time interval $0-5 \mathrm{sec}, 5-10 \mathrm{sec}$ and $10^{-15} \mathrm{sec}$. order comes to be zero, but graph is not a straight line.
Assuming $1^{\text {st }}$ order kinetics
$K=\frac{1}{t} \ln \frac{A_{0}}{A_{t}}$
$K=\frac{1}{10} \ln \frac{40}{20}$
Time required to reduce to $2.5 g / L$
$K=\frac{1}{t} \ln \frac{50}{2.5}$
$\frac{1}{10} \ln 2=\frac{1}{t} \ln 20$
$t=\frac{1.3010 \times 10}{0.3010}=43.3 \mathrm{sec}$.
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